Independent and Dependent Two-Sample t-Tests
STA4173 Lecture 3, Summer 2023
Independent vs. Dependent Data
Independent: An individual selected for one sample does not dictate which individual is to be in a second sample.
e.g., there are two sections of STA4173; sampling from each would be independent samples
- there is not a way for us to link the individuals in the samples
Dependent: An individual selected to be in one sample is used to determine the individual in the second sample.
e.g., sampling from one section of STA4173 and examining project grades over time would be dependent samples
- we can link on student ID
Independent vs. Dependent Data
Among competing acne medications, does one perform better than the other?
To answer this question, researchers applied Medication A to one part of the subject’s face and Medication B to a different part of the subject’s face to determine the proportion of subjects whose acne cleared up for each medication.
The part of the face that received Medication A was randomly determined.
Is this independent or dependent data?
Independent vs. Dependent Data
Among competing acne medications, does one perform better than the other?
To answer this question, researchers applied Medication A to one part of the subject’s face and Medication B to a different part of the subject’s face to determine the proportion of subjects whose acne cleared up for each medication.
The part of the face that received Medication A was randomly determined.
Is this independent or dependent data?
This is dependent data – the data can be linked by person.
Independent vs. Dependent Data
Do individuals who make fast-food purchases with a credit card tend to spend more than those who pay with cash?
To answer this question, a marketing manager randomly selects 30 credit-card receipts and 30 cash receipts to determine if the credit-card receipts have a significantly higher dollar amount, on average.
Is this independent or dependent data?
Independent vs. Dependent Data
Do individuals who make fast-food purchases with a credit card tend to spend more than those who pay with cash?
To answer this question, a marketing manager randomly selects 30 credit-card receipts and 30 cash receipts to determine if the credit-card receipts have a significantly higher dollar amount, on average.
Is this independent or dependent data?
This is independent data – the data cannot be linked.
Independent vs. Dependent Data
Are products purchased on Amazon less expensive than those purchased online at Walmart?
To answer this question, researchers randomly identified 20 products sold at both stores and determined the selling price at Amazon and the online Walmart store to determine if there was a significant difference in the price of the goods.
Is this independent or dependent data?
Independent vs. Dependent Data
Are products purchased on Amazon less expensive than those purchased online at Walmart?
To answer this question, researchers randomly identified 20 products sold at both stores and determined the selling price at Amazon and the online Walmart store to determine if there was a significant difference in the price of the goods.
Is this independent or dependent data?
This is dependent data – the data can be linked by item.
Estimating \mu_1-\mu_2
We are now interested in comparing two independent groups.
We assume that the two groups come from different populations where
\mu_i is the mean for group i,
\sigma^2_i is the standard deviation for group i, and
N_i is the population size of group i.
After drawing samples, we have the following,
\bar{x}_i estimates \mu_i,
s^2_i estimates \sigma^2_i, and
n_i is the sample size for group i.
Because we are interested in comparing groups, we are interested in \mu_1-\mu_2.
- We estimate this with \bar{x}_1-\bar{x}_2.
Confidence Interval for \mu_1-\mu_2
- \mathbf{(1-\boldsymbol\alpha)100\%} CI for the difference between two population means, \mathbf{\boldsymbol\mu_1-\boldsymbol\mu_2}:
(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2}}
where t_{\alpha/2} has \text{min}(n_1-1, n_2-1) degrees of freedom.
To construct this interval, we require either:
- the two populations to be normally distributed or
- the sample sizes are sufficiently large (n_1 \ge 30 and n_2 \ge 30)
Confidence Interval for \mu_1-\mu_2
- We will use the
t.test()function to find the CI,
Confidence Interval for \mu_1-\mu_2
In the Spacelab Life Sciences 2 payload, 14 male rats were sent to space.
Upon their return, the red blood cell mass (in milliliters) of the rats was determined.
A control group of 14 male rats was held under the same conditions (except for space flight) as the space rats, and their red blood cell mass was also determined when the space rats returned.
The project resulted in the following data:
8.59, 8.64, 7.43, 7.21, 6.39, 6.87, 7.89, 9.79, 6.85, 7.54, 7.00, 8.80, 9.30, 8.03
8.65, 6.99, 8.40, 9.66, 7.14, 7.62, 7.44, 8.55, 8.70, 9.14, 7.33, 8.58, 9.88, 9.94
- Construct the 99% confidence interval for the average difference in red blood cell mass.
Confidence Interval for \mu_1-\mu_2
library(tidyverse)
rbc <- c(8.59, 8.64, 7.43, 7.21, 6.39, 6.87, 7.89,
9.79, 6.85, 7.54, 7.00, 8.80, 9.30, 8.03,
8.65, 6.99, 8.40, 9.66, 7.14, 7.62, 7.44,
8.55, 8.70, 9.14, 7.33, 8.58, 9.88, 9.94) # enter blood cell mass
rat <- c(rep("Space",14), rep("Earth",14)) # enter identifier
data <- tibble(rat, rbc) # create dataset
t.test(rbc ~ rat, data = data, conf.level = 0.99)
Welch Two Sample t-test
data: rbc by rat
t = 1.4368, df = 25.996, p-value = 0.1627
alternative hypothesis: true difference in means between group Earth and group Space is not equal to 0
99 percent confidence interval:
-0.5130364 1.6116078
sample estimates:
mean in group Earth mean in group Space
8.430000 7.880714 - The 99% CI for \mu_\text{Earth}-\mu_\text{Space} is (-0.51, 1.61).
Confidence Interval for \mu_1-\mu_2
If the 99% CI for \mu_\text{Earth}-\mu_\text{Space} is (-0.51, 1.61),
Is there a difference in red blood cell mass between the space and earth rats?
Is the difference smaller than 2 ml?
Confidence Interval for \mu_1-\mu_2
If the 99% CI for \mu_\text{Earth}-\mu_\text{Space} is (-0.51, 1.61),
Is there a difference in red blood cell mass between the space and earth rats?
- No, 0 is contained in the interval.
Is the difference smaller than 2 ml?
- Yes, the interval is entirely below 2 and above -2.
Two-Sample t-Test
- Hypothesis Test for Two Independent Means:
Two-Sample t-Test
- We again use the
t.test()function to perform the test,
Important!!
We are estimating \mu_1 - \mu_2, but R is going to subtract in alphabetical or numeric order of the grouping variable.
e.g., if we have “Earth” and “Space”, it will estimate \mu_{\text{Earth}} - \mu_{\text{Space}}.
e.g., if we have “110” and “5”, it will estimate \mu_{5} - \mu_{110}.
In the case of two-tailed tets, this does not matter… but beware when doing a one-tailed test!
Two-Sample t-Test
- We again use the
t.test()function to perform the test,
- Recall the rat data,
What is the continuous variable?
What is the grouping variable?
Two-Sample t-Test
- We again use the
t.test()function to perform the test,
- Recall the rat data,
What is the continuous variable? rbc
What is the grouping variable? rat
Two-Sample t-Test
- Let’s now determine if there’s a difference in red blood cell mass between the earth and space rats. Test at the \alpha=0.01 level.
Welch Two Sample t-test
data: rbc by rat
t = 1.4368, df = 25.996, p-value = 0.1627
alternative hypothesis: true difference in means between group Earth and group Space is not equal to 0
95 percent confidence interval:
-0.2365544 1.3351258
sample estimates:
mean in group Earth mean in group Space
8.430000 7.880714 Two-Sample t-Test
Hypotheses
- H_0: \ \mu_{\text{Earth}} = \mu_{\text{Space}}
- H_1: \ \mu_{\text{Earth}} \ne \mu_{\text{Space}}
Test Statistic and p-Value
- t_0 = 1.437
- p = 0.163
Rejection Region
- Reject H_0 if p < \alpha; \alpha = 0.01.
Conclusion/Interpretation
Fail to reject H_0.
There is not sufficient evidence to suggest that there is a difference in the red blood cell mass between earth and space rats.
Example
- An experiment was conducted to evaluate the effectiveness of a treatment for tapeworm in the stomachs of sheep. A random sample of 24 worm-infested lambs of approximately the same age and health was randomly divided into two groups. Twelve of the lambs were injected with the drug and the remaining 12 were left untreated. After a 6-month period, the lambs were slaughtered, and the worm counts recorded as listed below.
18, 43, 28, 50, 16, 32, 13, 35, 38, 33, 6, 7
40, 54, 26, 63, 21, 37, 39, 23, 48, 58, 28, 39
- Is there significant evidence that the untreated lambs have a mean tapeworm count that is more than five units greater than the mean count for the treated lambs? Use \alpha=0.10.
Example
- Let’s first create the data.
Example
- Let’s now summarize the data.
One Sample t-test
data: data$worms
t = 10.582, df = 23, p-value = 2.6e-10
alternative hypothesis: true mean is not equal to 0
90 percent confidence interval:
27.76022 38.48978
sample estimates:
mean of x
33.125 - Overall \bar{y}=33.125 with a 90% CI for \mu of (27.76, 38.49).
Example
- Further summarizing at the group level, for treated
worms_trt <- data %>% filter(trt == "Treated") # only treated sheep
t.test(worms_trt$worms)[5] # mean only$estimate
mean of x
26.58333 $conf.int
[1] 19.13771 34.02895
attr(,"conf.level")
[1] 0.9- \bar{y}_{\text{T}} = 26.58, 90% CI for \mu_{\text{T}} (19.14, 34.03)
Example
- Further summarizing at the group level, for untreated
worms_untrt <- data %>% filter(trt == "Not") # only untreated sheep
t.test(worms_untrt$worms)[5] # mean only$estimate
mean of x
39.66667 $conf.int
[1] 32.48199 46.85134
attr(,"conf.level")
[1] 0.9- \bar{y}_{\text{U}} = 39.67, 90% CI for \mu_{\text{U}} (32.48, 46.86)
Example
Putting our descriptives together:
- Overall \bar{y}=33.125, 90% CI for \mu (26.66, 39.60).
- \bar{y}_{\text{T}} = 26.58, 90% CI for \mu_{\text{T}} (19.14, 34.03)
- \bar{y}_{\text{U}} = 39.67, 90% CI for \mu_{\text{U}} (32.48, 46.86)
What do we hypothesize will happen when we look at inference on \mu_{\text{T}} - \mu_{\text{U}}?
Example
- Let’s first look at the 90% CI for \mu_{\text{T}} - \mu_{\text{U}},
Welch Two Sample t-test
data: worms by trt
t = 2.2709, df = 21.972, p-value = 0.03331
alternative hypothesis: true difference in means between group Not and group Treated is not equal to 0
90 percent confidence interval:
3.189613 22.977054
sample estimates:
mean in group Not mean in group Treated
39.66667 26.58333 90% CI for \mu_{\text{U}} - \mu_{\text{T}} is (3.19, 22.98).
Is there significant evidence that the untreated lambs have a mean tapeworm count that is more than five units greater than the mean count for the treated lambs?
Example
- Let’s first look at the 90% CI for \mu_{\text{U}} - \mu_{\text{T}},
Welch Two Sample t-test
data: worms by trt
t = 2.2709, df = 21.972, p-value = 0.03331
alternative hypothesis: true difference in means between group Not and group Treated is not equal to 0
90 percent confidence interval:
3.189613 22.977054
sample estimates:
mean in group Not mean in group Treated
39.66667 26.58333 90% CI for \mu_{\text{U}} - \mu_{\text{T}} is (3.19, 22.98).
Is there significant evidence that the untreated lambs have a mean tapeworm count that is more than five units greater than the mean count for the treated lambs?
- No, the CI contains 5……
- …. but this is a two-sided CI and we are performing a one-tailed test
Example
Let’s now look at the formal hypothesis test.
Is there significant evidence that the untreated lambs have a mean tapeworm count that is more than five units greater than the mean count for the treated lambs?
- This translates to \mu_{\text{U}} - \mu_{\text{T}} > 5 for the alternative.
Welch Two Sample t-test
data: worms by trt
t = 1.403, df = 21.972, p-value = 0.08729
alternative hypothesis: true difference in means between group Not and group Treated is greater than 5
90 percent confidence interval:
5.470851 Inf
sample estimates:
mean in group Not mean in group Treated
39.66667 26.58333 Example
Hypotheses
- H_0: \ \mu_{\text{Untreated}} - \mu_{\text{Treated}} \le 5
- H_1: \ \mu_{\text{Untreated}} - \mu_{\text{Treated}} > 5
Test Statistic and p-Value
- t_0 = 21.97
- p = 0.087
Rejection Region
- Reject H_0 if p < \alpha; \alpha = 0.10.
Conclusion/Interpretation
Reject H_0.
There is sufficient evidence to suggest that untreated lambs have a mean tapeworm count that is more than five units greater than the mean count for the treated lambs.
Example
Let’s compare the two CIs for \mu_{\text{U}} - \mu_{\text{T}},
- Two-sided 90% CI for \mu_{\text{U}} - \mu_{\text{T}} is (3.19, 22.98).
- One-sided 90% CI for \mu_{\text{U}} - \mu_{\text{T}} is (5.47, \infty).
Note that when we are “close to the boundary” for one-tailed hypothesis testing, our two-sided CIs may not match the hypothesis test results.
- We can tell that we are “close to the boundary” because p = 0.087 and the lower limit of the one-sided CI is 5.47.
Independent vs. Dependent Data
Independent: An individual selected for one sample does not dictate which individual is to be in a second sample.
e.g., there are two sections of STA4173; sampling from each would be independent samples
- there is not a way for us to link the individuals in the samples
Dependent: An individual selected to be in one sample is used to determine the individual in the second sample.
e.g., sampling from one section of STA4173 and examining project grades over time would be dependent samples
- we can link on student ID
Estimating \mu_1-\mu_2
We are now interested in comparing two dependent groups.
We assume that the two groups come from the same population and are going to examine the difference,
d = y_{i, 1} - y_{i, 2}
After drawing samples, we have the following,
\bar{d} estimates \mu_d,
s^2_d estimates \sigma^2_d, and
n is the sample size.
Confidence Interval for \mu_d
- \mathbf{(1-\boldsymbol\alpha)100\%} CI for the difference between two population means, \mathbf{\boldsymbol\mu_d}:
\bar{d} \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}
where t_{\alpha/2} has n-1 degrees of freedom.
To construct this interval, we require either:
- the differences to be normally distributed or
- the sample size is sufficiently large (n \ge 30)
Confidence Interval for \mu_d
- We will use the
t.test()function to find the CI,
Confidence Interval for \mu_d
Insurance adjusters are concerned about the high estimates they are receiving for auto repairs from garage I compared to garage II.
15 cars were taken to both garages for separate estimates of repair costs.
- Construct the 95% confidence interval for the average difference between the two garages.
Confidence Interval for \mu_d
Paired t-test
data: garage$g1 and garage$g2
t = 6.0234, df = 14, p-value = 3.126e-05
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
0.3949412 0.8317254
sample estimates:
mean difference
0.6133333 - The 95% CI for \mu_d, where d = x_{\text{I}} - x_{\text{II}} is (0.39, 0.83).
Confidence Interval for \mu_d
From the problem statement:
- Insurance adjusters are concerned about the high estimates they are receiving for auto repairs from garage I compared to garage II.
Our CI is (0.39, 0.83) – can we say that estimates from garage I are higher than those from garage II?
Confidence Interval for \mu_d
From the problem statement:
- Insurance adjusters are concerned about the high estimates they are receiving for auto repairs from garage I compared to garage II.
Our CI is (0.39, 0.83) – can we say that estimates from garage I are higher than those from garage II?
- Yes; the CI is entirely above 0 and the subtraction order was garage I - garage II.
Paired t-Test
- Hypothesis Test for Two Dependent Means:
Paired t-Test
- We again use the
t.test()function to perform the test,
Important!!
- R is going to subtract in the order that we plugged var1 and var2 into the
t.test().
- R is going to subtract in the order that we plugged var1 and var2 into the
Paired t-Test
- Let’s now formally determine if garage I’s estimates are higher than garage II’s. Test at the \alpha=0.05 level.
Paired t-test
data: garage$g1 and garage$g2
t = 6.0234, df = 14, p-value = 1.563e-05
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
0.4339886 Inf
sample estimates:
mean difference
0.6133333 Paired t-Test
Hypotheses
- H_0: \ \mu_{\text{I}} \le \mu_{\text{II}} OR \mu_{d} \le 0, where \mu_d = \mu_{\text{I}} - \mu_{\text{II}}
- H_1: \ \mu_{\text{I}} > \mu_{\text{II}} OR \mu_{d} > 0
Test Statistic and p-Value
- t_0 = 6.023
- p < 0.001
Rejection Region
- Reject H_0 if p < \alpha; \alpha = 0.05.
Conclusion/Interpretation
Reject H_0.
There is sufficient evidence to suggest the estimates at garage I are higher than that of garage II.
Example
Professor Neill measured the time (in seconds) required to catch a falling meter stick for 12 randomly selected students’ dominant and nondominant hands.
A coin flip is used to determine whether reaction time is measured using the dominant or nondominant hand first.
Professor Neill wants to know if the reaction time in an individual’s dominant hand is equal to the reaction time in their nondominant hand.
First, find the 99% CI for the difference in reaction times.
Then, formally test to determine if there is a difference; test at the \alpha=0.01 level.
Example
Let’s first describe the data.
Our first step will be to find the difference so that we can look at \bar{d} and s_d.
- Is the reaction time in an individual’s dominant hand equal to the reaction time in their nondominant hand?
Example
- We would like to know what the distribution looks like, so let’s create a box plot,
Example
- We would like to know what the distribution looks like, so let’s create a box plot,
Example
- To describe, we will look at means and standard deviations:
students %>% summarize(mean(dom), sd(dom), # summary of dominant hand
mean(non), sd(non), # summary of non-dominant hand
mean(d), sd(d)) # summary of difference| Mean (Std. Dev) | |
|---|---|
| Dominant Hand | 0.180 (0.018) |
| Non-dominant Hand | 0.193 (0.018) |
| Difference (Dominant - Non-dominant) | -0.013 (0.016) |
Example
- Let’s now find the 99% CI for the difference in reaction time (\mu_d).
Paired t-test
data: students$dom and students$non
t = -2.7759, df = 11, p-value = 0.01803
alternative hypothesis: true mean difference is not equal to 0
99 percent confidence interval:
-0.02789797 0.00156464
sample estimates:
mean difference
-0.01316667 - The 99% CI for \mu_d, where d = x_{\text{dominant}} - x_{\text{non-dominant}} is (-0.028, 0.002).
Example
The 99% CI for \mu_d, where d = x_{\text{dominant}} - x_{\text{non-dominant}} is (-0.028, 0.002).
- Can we say that there is a difference in the reaction times?
Example
The 99% CI for \mu_d, where d = x_{\text{dominant}} - x_{\text{non-dominant}} is (-0.028, 0.002).
Can we say that there is a difference in the reaction times?
- No, 0 is included in the interval
Example
- Let’s now formally determine there is a difference between the reaction of dominant and non-dominant hands.
Paired t-test
data: students$dom and students$non
t = -2.7759, df = 11, p-value = 0.01803
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
-0.023606270 -0.002727063
sample estimates:
mean difference
-0.01316667 Example
Hypotheses
- H_0: \ \mu_{\text{dominant}} = \mu_{\text{non-dominant}} OR \mu_{d} = 0, where \mu_d = \mu_{\text{dominant}} - \mu_{\text{non-dominant}}
- H_1: \ \mu_{\text{dominant}} \ne \mu_{\text{non-dominant}} OR \mu_{d} \ne 0
Test Statistic and p-Value
- t_0 = -2.2776
- p = 0.018
Rejection Region
- Reject H_0 if p < \alpha; \alpha = 0.01.
Conclusion/Interpretation
Fail to reject H_0.
There is not sufficient evidence to suggest that there is a difference in reaction times between dominant and non-dominant hands.